Question: Solve for $t$, $ \dfrac{10}{9t + 12} = -\dfrac{1}{6t + 8} + \dfrac{2t + 1}{15t + 20} $
Explanation: First we need to find a common denominator for all the expressions. This means finding the least common multiple of $9t + 12$ $6t + 8$ and $15t + 20$ The common denominator is $90t + 120$ To get $90t + 120$ in the denominator of the first term, multiply it by $\frac{10}{10}$ $ \dfrac{10}{9t + 12} \times \dfrac{10}{10} = \dfrac{100}{90t + 120} $ To get $90t + 120$ in the denominator of the second term, multiply it by $\frac{15}{15}$ $ -\dfrac{1}{6t + 8} \times \dfrac{15}{15} = -\dfrac{15}{90t + 120} $ To get $90t + 120$ in the denominator of the third term, multiply it by $\frac{6}{6}$ $ \dfrac{2t + 1}{15t + 20} \times \dfrac{6}{6} = \dfrac{12t + 6}{90t + 120} $ This give us: $ \dfrac{100}{90t + 120} = -\dfrac{15}{90t + 120} + \dfrac{12t + 6}{90t + 120} $ If we multiply both sides of the equation by $90t + 120$ , we get: $ 100 = -15 + 12t + 6$ $ 100 = 12t - 9$ $ 109 = 12t $ $ t = \dfrac{109}{12}$